A complex number is *algebraic* if it is the root of some polynomial with rational coefficients. is algebraic (e.g. the polynomial ); is algebraic (e.g. the polynomial ); and are not. A complex number that is not algebraic is called *transcendental*.

Is the sum of algebraic numbers always algebraic? What about the product of algebraic numbers? For example, given that and are algebraic, how do we know is also algebraic?

We can try to repeatedly square the equation . This gives us . Then isolating the radical, we have . Squaring again, we get , so is a root of . This is in fact the unique monic polynomial of minimum degree that has as a root (called the *minimal polynomial* of ) which shows is algebraic. But a sum like would probably have a minimal polynomial of much greater degree, and it would be much more complicated to construct this polynomial and verify the number is algebraic.

It is also increasingly difficult to construct polynomials for numbers like . And, apparently there also exist algebraic numbers that cannot be expressed as radicals at all. This further complicates the problem.

In fact, the sum and product of algebraic numbers *is* algebraic – in fact, any number which can be obtained by adding, subtracting, multiplying and dividing algebraic numbers is algebraic. This means that a number like

is algebraic – there is *some* polynomial which has it as a root. But we will prove this result *non-constructively;* that is, we will prove that such a number must be algebraic, without providing an explicit process to actually obtain the polynomial. To establish this result, we will try to look for a deeper structure in the algebraic numbers, which is elucidated, perhaps surprisingly, using the tools of linear algebra.

**Definition**: Let be a set of complex numbers that contains . We say is an *abelian group* if for all , and . In other words, is closed under addition and subtraction.

Some examples: , , , , the Gaussian integers (i.e. the set of expressions where and are integers)

**Definition**: An abelian group is a *field* if it contains and for all , , and if , . In other words, is closed under multiplication and division.

We generally use the letters , , for arbitrary fields. Of the previous examples, only , , are fields.

**Exercise**: Show that if is a field, .

**Exercise**: Show that the set is a field. We call this field . (Hint: rationalize the denominator).

**Exercise**: Describe the the smallest field which contains both and (this is called )

Generally if is a field and some complex numbers, we will denote the smallest field that contains all of and the elements as .

**Definition**: Let be a field. A *-vector space* is an abelian group such that if , , then . Intuitively, the elements of are closed under *scaling* by . (We also sometimes use the phrase “ is a vector space over “)

Examples: and are both -vector spaces. In fact, is also a vector space.

With this language in place, we can state the main goal of this post as follows:

**Theorem**: If are algebraic numbers and , then is algebraic.

Note how this encompasses our previous claim that any number obtained by adding, subtracting, multiplying and dividing algebraic numbers is algebraic. For example, we can deduce the giant fraction above is algebraic by applying the theorem to .

**Definition**: A field extension of a field is just a field that contains . We will write this as “ is a field extension”.

**Exercise**: If is a field extension, then is a -vector space

Consider a field like . Not only is it a field, but it is also a -vector space – its elements can be scaled by rational numbers. We want to adapt concepts from ordinary linear algebra to this setting. We want to think of as a two dimensional vector space, with basis and – every element is can be uniquely written as a -linear combination of and . Similarly, we would like to think of as a -vector space with basis . Note that if we regard as a -vector space, its basis is . This is because if an element is written as , where the coefficients are rational, we can rewrite it as , now with coefficients in .

On the other hand, consider . Since is not algebraic, no linear combination (with rational coefficients) of the numbers equals without all the coefficients being zero. This makes them *linearly independent* – and makes an infinite-dimensional vector space.

This finite-versus-infinite dimensionality difference will lie at the crux of our argument. Here is a rough summary of the argument to prove the Theorem:

- If we add an algebraic number to a field, we get a finite-dimensional vector space over that field.
- By repeatedly adding algebraic numbers to , the resulting field will still be a finite-dimensional -vector space.
- Any element of a field which is a finite-dimensional -vector space must be algebraic.

From here on, we will assume that the reader has some familiarity with ideas from linear algebra, such as the notions of linear combination, linear independence, and basis. We will only sketch proofs (maybe I will add details later). We would like to warn the reader that the proof sketches have many gaps and may be difficult to follow.

**Definition**: A field extension is *finite* if there is a finite set of elements such that every element of can be represented as , where all the . We say the elements *generate* (aka span) over . If, furthermore, every element can be represented uniquely in this form, then we say is a –*basis* (or just basis) for .

Examples: and are finite extensions, while is not.

**Exercise**: Show is not a finite extension. (Hint: is uncountable)

**Proposition 1**: Every finite field extension has a basis.

*Proof sketch:* This is a special case of a famous theorem in linear algebra. Assume generate . Start with the last element. If can be represented as a linear combination of , remove it from the list. If we removed , we now have , and we can repeat the process with . Continue this process until we cannot remove any more elements. Then (it can be checked that) we obtain a set whose elements are linearly independent. Then (it can be checked that) these form a -basis for .

For the next theorem, keep in mind the example of the extensions and .

**Proposition 2**: Suppose and are finite field extensions. Then is a finite extension.

*Proof*: By Proposition 1, both these field extensions have bases. Label the -basis of as , and the -basis for as . Then every element of can be uniquely represented as , for . Furthermore, each can be written uniquely as for . Plugging these in for the shows that the elements of the form form a -basis for .

**Proposition 3**: If is algebraic, then is a finite extension.

*Proof sketch*: Suppose . First we will show every element in is a polynomial in with coefficients in . consists of all numbers that can be obtained by repeatedly adding, subtracting, multiplying and dividing and elements of . By performing some algebraic manipulations and combining fractions, we can see that every element can be written in the form , where and are polynomials with coefficients in .

Now I claim that for any polynomial where , there exists a polynomial such that . Let be the minimal polynomial of . Since and , and are relatively prime as polynomials. So there exist polynomials and such that . Plugging in into the polynomials gives us . Since , , as desired.

Thus any element in , written as , can be written as for an appropriate polynomial ; in other words, every element of is a polynomial of with coefficients in . Now, suppose the minimal polynomial of is . Then . So (and all higher powers of ) can be written as rational linear combinations of . Then it can be verified that every element in can be written uniquely as a -linear combination of . This establishes that is a finite extension; in particular, it has the basis

**Proposition 4**: Suppose is a finite extension. Then any is algebraic.

*Proof sketch*: Consider the elements . They cannot all be linearly independent. This is because in a finite-dimensional vector space of dimension any set with at least elements must be linearly dependent. So there must be some linear combination of them which equals zero. But this simply means that for some constants and positive integer , . Letting the be the coefficients of a polynomial , , so is algebraic.

We have developed enough theory now to prove the result.

*Proof of main Theorem*: Let be some collection of algebraic numbers. Then by Proposition 3, are all finite field extensions. Applying Proposition 2 repeatedly gives us is a finite extension. Then by Proposition 4, any element of is algebraic.

This theorem essentially says that given a collection of algebraic numbers, by repeatedly adding, subtracting, multiplying and dividing them, we can only obtain algebraic numbers. But what about radicals? For example, we have now established that is algebraic. Is algebraic as well?

In fact, we can generalize our result to the following: any number obtained by repeatedly adding, subtracting, multiplying, dividing algebraic numbers, as well as taking -th roots (for positive integers ) will be algebraic.

This is not too hard now that we have the main theorem. It suffices to show that if is algebraic, is algebraic. If is algebraic, then there exist constants such that . This can be rewritten as . Thus is algebraic.

*Acknowledgements*: Thanks to Anton Cao and Nagaganesh Jaladanki for reviewing this article.